Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {
	static class fastreader {
		BufferedReader br;
		StringTokenizer st;

		public fastreader() {
			BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
			this.br = br;
		}

		String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException ie) {
				}
			}
			return st.nextToken().toString();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}
	}

	public static void main(String[] args) {

		fastreader fr = new fastreader();
		int n = fr.nextInt();
		int m = fr.nextInt();
		int ar[][] = new int[n + 1][m + 1];
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= m; j++) {
				ar[i][j] = fr.nextInt();
			}
		}
int count=0;
		StringBuilder sb = new StringBuilder();
		if (n < m) {
			int rowmin[] = new int[n + 1];
			for (int i = 0; i <= n; i++)
				rowmin[i] = 1000000;
			for (int i = 1; i <= n; i++) {
				for (int j = 1; j <= m; j++) {
					rowmin[i] = Math.min(rowmin[i], ar[i][j]);
				}
			}
			for (int i = 1; i <= n; i++) {
				if (rowmin[i] != 0) {
					for (int x = 1; x <= rowmin[i]; x++)
						{sb.append("row " + i + "\n");count++;}
				}
			}
			for (int i = 1; i <= n; i++) {
				for (int j = 1; j <= m; j++) {
					ar[i][j] = ar[i][j] - rowmin[i];
				}
			}

			int colmin[] = new int[m + 1];
			for (int i = 0; i <= m; i++)
				colmin[i] = 1000000;
			for (int i = 1; i <= m; i++) {
				for (int j = 1; j <= n; j++) {
					colmin[i] = Math.min(colmin[i], ar[j][i]);
				}
			}
			for (int i = 1; i <= m; i++) {
				if (colmin[i] != 0) {
					for (int x = 1; x <= colmin[i]; x++)
						{sb.append("col " + i + "\n");count++;}
				}
			}
			for (int i = 1; i <= m; i++) {
				for (int j = 1; j <= n; j++) {
					ar[j][i] = ar[j][i] - colmin[i];
				}
			}
			boolean flag = true;
			for (int i = 1; i <= n; i++){
				for (int j = 1; j <= m; j++) {
					if (ar[i][j] != 0) {
						flag = false;
						break;
					}
				}
				if(flag==false)break;
				}
if(flag==false){System.out.println(-1);}
else{System.out.println(count);System.out.println(sb);}
		}
		else
		{
			int colmin[] = new int[m + 1];
			for (int i = 0; i <= m; i++)
				colmin[i] = 1000000;
			for (int i = 1; i <= m; i++) {
				for (int j = 1; j <= n; j++) {
					colmin[i] = Math.min(colmin[i], ar[j][i]);
				}
			}
			for (int i = 1; i <= m; i++) {
				if (colmin[i] != 0) {
					for (int x = 1; x <= colmin[i]; x++)
						{sb.append("col " + i + "\n");count++;}
				}
			}
			for (int i = 1; i <= m; i++) {
				for (int j = 1; j <= n; j++) {
					ar[j][i] = ar[j][i] - colmin[i];
				}
			}
			
			
			int rowmin[] = new int[n + 1];
			for (int i = 0; i <= n; i++)
				rowmin[i] = 1000000;
			for (int i = 1; i <= n; i++) {
				for (int j = 1; j <= m; j++) {
					rowmin[i] = Math.min(rowmin[i], ar[i][j]);
				}
			}
			for (int i = 1; i <= n; i++) {
				if (rowmin[i] != 0) {
					for (int x = 1; x <= rowmin[i]; x++)
						{sb.append("row " + i + "\n");count++;}
				}
			}
			for (int i = 1; i <= n; i++) {
				for (int j = 1; j <= m; j++) {
					ar[i][j] = ar[i][j] - rowmin[i];
				}
			}

		
			boolean flag = true;
			for (int i = 1; i <= n; i++){
				for (int j = 1; j <= m; j++) {
					if (ar[i][j] != 0) {
						flag = false;
						break;
					}
				}
				if(flag==false)break;
				}
if(flag==false){System.out.println(-1);}
else{System.out.println(count);System.out.println(sb);}
		}
		
		
		

	}

}